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# A certain company assigns employees to offices in such a way

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Senior Manager
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A certain company assigns employees to offices in such a way [#permalink]

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29 Apr 2006, 21:47
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A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9
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30 Apr 2006, 06:00
Since the employees are distinguishabe it should be 3^2 = 9

if the same question is asked about oranges or apples it is (3+2-1)C(2-1) = 4 (since empty groups are allowed)

If empty groups aint allowed for apples or oranges it is (3-1)C(2-1) = 2

Correct me guys if I am wrong
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30 Apr 2006, 07:49
trivikram wrote:
Since the employees are distinguishabe it should be 3^2 = 9

if the same question is asked about oranges or apples it is (3+2-1)C(2-1) = 4 (since empty groups are allowed)

If empty groups aint allowed for apples or oranges it is (3-1)C(2-1) = 2

Correct me guys if I am wrong

Explain how you got (3+2-1)C(2-1) and (3-1)C(2-1)
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30 Apr 2006, 07:56
Each employee can be assigned to one of 2 offices
hence the total amount of possibilities is 2 * 2 * 2 = 8
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30 Apr 2006, 09:47
deowl wrote:
Each employee can be assigned to one of 2 offices
hence the total amount of possibilities is 2 * 2 * 2 = 8

Very clever! I counted all possibilities and got 8
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01 May 2006, 01:33
deowl wrote:
Each employee can be assigned to one of 2 offices
hence the total amount of possibilities is 2 * 2 * 2 = 8

Yes the OA is 8,
Do you have another approach to solve this question?
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01 May 2006, 01:53
yes I have
we have 4 distinct possibilities for employees assignments

3 0 ( 1 way )
0 3 ( 1 way )
1 2 ( 3 ways - choose the one for the first room , while the rest go to the second )
2 1 ( 3 ways - same as above )

so 1 + 1 + 3 + 3 = 8

however the previous solution is much simplier as you can see
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01 May 2006, 02:16
deowl wrote:
yes I have
we have 4 distinct possibilities for employees assignments

3 0 ( 1 way )
0 3 ( 1 way )
1 2 ( 3 ways - choose the one for the first room , while the rest go to the second )
2 1 ( 3 ways - same as above )

so 1 + 1 + 3 + 3 = 8

however the previous solution is much simplier as you can see

Thank you deowl, but what do you mean that one is simplest?
Please elaborate it 2*2*2 what is 2 in this case? thank you.
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01 May 2006, 03:45
2 is the number of ways each one can be assigned ( one of 2 rooms )
so it is 2^3 for 3 persons
[#permalink] 01 May 2006, 03:45
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# A certain company assigns employees to offices in such a way

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